Miguel took 7 hours to read a 259-page book and Julissa took 5 hours to read a 175-page book. Who read more pages per hour?

Mathematics

1 answer:

Pie2 years ago

6 0

M: 259/7=37 pages per hour
J: 175/5= 35 pages per hour

Answer: Miguel

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Help on area of composite figures

balandron [24]

break it up into simpler shapes. See the photo.
There is:
trapezoid
A = 0.5(3.2+6.5)4.5
= 21.825

triangle
A = 0.5(2.7)4.5
= 6.075

rectangle
A = 1.8(4.5)
= 8.1

Total Area = 21.825 + 6.075 + 8.1
= 36

7 0

3 years ago

Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b

trapecia [35]

Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

$6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) $\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}$

( 2 ) $\sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}$

These two identities makes sin(π / 10) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and cos(π / 10) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ .

Therefore cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ . Substitute,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$

And now simplify this expression to receive our answer,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$ = $-\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i$ ,

$-\frac{3\sqrt{5+\sqrt{5}}}{4}$ = $-2.01749\dots$ and $\:\frac{3\sqrt{3-\sqrt{5}}}{4}$ = $0.65552\dots$

= $-2.01749+0.65552i$

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ , cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

$6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}$